Binomial Expansions: Complete Theory and Formulas 1
Binomial Expansions: Complete Theory and Formulas
1. Introduction to Binomial Coefficients
Binomial coefficients are fundamental mathematical concepts used in algebra and probability. They are represented in two ways:
• (n r) or (n choose r)
2. Methods to Calculate Binomial Coefficients
2.1 Pascal's Triangle Method
Pascal's Triangle provides a visual way to find binomial coefficients:
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
2.2 Formula Method
or
(n r) = n × (n-1) × (n-2) × ... × (n-r+1) / (r × (r-1) × (r-2) × ... × 1)
3. The Binomial Theorem
For any positive integer n, the expansion of (1 + x)ⁿ is given by:
4. General Form of Binomial Expansion
For any constants a and b, and positive integer n:
5. Alternative Form
The expansion of (1 + x)ⁿ can also be written as:
6. Key Properties
- The number of terms in the expansion is (n + 1)
- The coefficients are symmetric
- The sum of coefficients equals 2ⁿ
7. Example Application
Expand (1 + x)⁴:
(1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴
8. Practice Problems
- Find the coefficient of x³ in (2 + x)⁵
- Expand (1 - x)³
- Calculate ⁵C₃ using both methods
Binomial Expansion Practice Problems
Master Binomial Theorem with These Practice Questions
A. Basic Binomial Expansion
Question 1 Easy
Expand (1 + x)³ using the binomial theorem.
Solution:
Let's solve this step by step:
1. Using the formula (1 + x)ⁿ where n = 3
2. The terms will be:
- (³C₀)x⁰ = 1
- (³C₁)x¹ = 3x
- (³C₂)x² = 3x²
- (³C₃)x³ = x³
Answer: 1 + 3x + 3x² + x³
Question 2 Medium
Find the coefficient of x⁴ in the expansion of (2 - x)⁶
Solution:
1. Using (a + b)ⁿ formula where a = 2, b = -x, n = 6
2. The general term is (⁶Cᵣ)(2)⁶⁻ʳ(-x)ʳ
3. For x⁴, r = 4
4. Therefore:
- Coefficient = ⁶C₄ × 2² × (-1)⁴
- = 15 × 4 × 1
- = 60
Answer: 60
Question 3 Hard
In the expansion of (1 + ax)ⁿ, the coefficients of the second and third terms are 15 and 90 respectively. Find the values of n and a.
Solution:
1. Second term coefficient: nC₁ × a = 15
2. Third term coefficient: nC₂ × a² = 90
3. From first equation: n × a = 15
4. From second equation: n(n-1)a²/2 = 90
5. Substituting n = 15/a in second equation:
6. (15/a)(14/a)a²/2 = 90
7. 105 = 90
8. Therefore: n = 6 and a = 2.5
Answer: n = 6, a = 2.5
Question 4 Hard
Find the middle term in the expansion of (2x - 1/x)⁸
Solution:
1. In an expansion of (ax + b)ⁿ, middle term occurs when powers of x are equal
2. Here, (2x)ʳ × (−1/x)⁸⁻ʳ should give x⁰
3. Therefore: r - (8-r) = 0
4. 2r - 8 = 0
5. r = 4
6. Middle term = ⁸C₄ × 2⁴ × (-1/x)⁴
7. = 70 × 16 × 1 = 1120
Answer: 1120
B. Advanced Binomial Applications
Question 5 Medium
In the expansion of (1 + 2x)⁷, find the sum of coefficients of terms where the power of x is even.
Solution:
1. The terms with even powers of x are: x⁰, x², x⁴, x⁶
2. The coefficients are:
- For x⁰: ⁷C₀ = 1
- For x²: ⁷C₂ × 2² = 21 × 4 = 84
- For x⁴: ⁷C₄ × 2⁴ = 35 × 16 = 560
- For x⁶: ⁷C₆ × 2⁶ = 7 × 64 = 448
3. Sum = 1 + 84 + 560 + 448 = 1093
Answer: 1093
Question 6 Hard
Find the term independent of x in the expansion of (x + 1/x)⁶
Solution:
1. For a term to be independent of x, the powers of x must sum to zero
2. In general term: ⁶Cᵣ × xʳ × (1/x)⁶⁻ʳ = ⁶Cᵣ × xʳ⁻⁶⁺ʳ = ⁶Cᵣ × x²ʳ⁻⁶
3. For term to be independent of x: 2r - 6 = 0
4. Therefore: r = 3
5. The coefficient will be ⁶C₃ = 20
Answer: 20
Note: This is equivalent to finding the coefficient of x⁰ in the expansion.
Question 7 Hard
The coefficient of x³ in (1 + ax)⁵ is equal to the coefficient of x² in (1 + ax)⁴. Find the value of a.
Solution:
1. For (1 + ax)⁵, coefficient of x³ = ⁵C₃ × a³
2. For (1 + ax)⁴, coefficient of x² = ⁴C₂ × a²
3. Therefore:
- ⁵C₃ × a³ = ⁴C₂ × a²
- 10 × a³ = 6 × a²
- 10a³ - 6a² = 0
- a²(10a - 6) = 0
- a = 0 or a = 3/5
4. Since a ≠ 0 (as it would make all terms zero)
Answer: a = 3/5
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